Chapter 19, Problem 91GQ

(a)

Interpretation Introduction

Interpretation:

The ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{0}}$ for the following reaction has to be determined.

${\text{(a) 3Cu(s) + 2NO}}_{\text{3}}^{\text{-}}{\text{(aq) + 8H}}^{\text{+}}\text{(aq) }\to {\text{3Cu}}^{\text{2+}}{\text{(aq) + 2NO(g) + 4H}}_{\text{2}}\text{O(l)}$.

Concept introduction:

According to the first law of thermodynamics, the change in internal energy of a system is equal ti the heat added to the sysytem minus the work done by the system.

The equation is as follows.

$\begin{array}{c}\text{ΔU = Q - W}\\ \text{ΔU = Change in internal energy}\\ \text{Q = Heat added to the system}\\ \text{W=Work done by the system}\end{array}$

In voltaic cell, the maximum cell potential is directly related to the free energy difference between the reactants and products in the cell.

$\begin{array}{c}{\text{ΔG}}^{\text{0}}{\text{= -nFE}}^{\text{0}}\\ \text{n = Number of moles transferred per mole of reactant and products}\\ \text{F = Faradayconstant=96485C/mol}\\ {\text{ E}}^{\text{0}}\text{= Volts = Work(J)/Charge(C)}\end{array}$

The relation between standard cell potential and equilibrium constant is as follows.

$\text{lnK = }\frac{{\text{nE}}^{\text{0}}}{\text{0}\text{.0257}}\text{ at 298K}$

(b)

Interpretation Introduction

Interpretation:

The ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{0}}$ for the following reaction has to be determined.

(b) ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}{\text{(aq) + 2Cl}}^{\text{-}}{\text{(aq) + 2H}}^{\text{+}}\text{(aq) }\to {\text{Cl}}_{\text{2}}{\text{(g) + 2H}}_{\text{2}}\text{O(l)}$.

Concept introduction:

According to the first law of thermodynamics, the change in internal energy of a system is equal ti the heat added to the sysytem minus the work done by the system.

The equation is as follows.

$\begin{array}{c}\text{ΔU = Q - W}\\ \text{ΔU = Change in internal energy}\\ \text{Q = Heat added to the system}\\ \text{W=Work done by the system}\end{array}$

In voltaic cell, the maximum cell potential is directly related to the free energy difference between the reactants and products in the cell.

$\begin{array}{c}{\text{ΔG}}^{\text{0}}{\text{= -nFE}}^{\text{0}}\\ \text{n = Number of moles transferred per mole of reactant and products}\\ \text{F = Faradayconstant=96485C/mol}\\ {\text{ E}}^{\text{0}}\text{= Volts = Work(J)/Charge(C)}\end{array}$

The relation between standard cell potential and equilibrium constant is as follows.

$\text{lnK = }\frac{{\text{nE}}^{\text{0}}}{\text{0}\text{.0257}}\text{ at 298K}$